Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a Wiener process $W_t$ which is adapted to $\mathscr{F}_t$, where this filtration has all of the standard properties. I'm also working with a stock-standard probability space here.

I want to know if the following useful identities are correct:

  • $W_t = {1}_{\{W_t \geq 0\}}W_t + {1}_{\{W_t < 0\}}W_t$

  • $|W_t| = {1}_{\{W_t \geq 0\}}W_t - {1}_{\{W_t < 0\}}W_t$

Note that I mean "$=$" as actually equal and not only equal in distribution.

share|cite|improve this question
up vote 1 down vote accepted

Hint: For every real number $x$, $$\mathbf 1_{\{x \geqslant 0\}} + \mathbf 1_{\{x < 0\}}=1,\qquad x\mathbf 1_{\{x \geqslant 0\}}- x\mathbf 1_{\{x < 0\}}= |x| $$

share|cite|improve this answer
So this confirms the two identifies path-wise. (i.e. for $W_t(\omega)$ and $|W_t(\omega)|$). However, I'm not quite sure how this can then be extended to demonstrate the case for the full $W_t$ random variable? – Jase Nov 8 '12 at 17:08
What you call the case for the full random variable is not clear to me. Random variables $X,Y:\Omega\to\mathbb R$ are such that $X=Y$ (almost surely) if and only if $X(\omega)=Y(\omega)$ for (almost) every $\omega$ in $\Omega$. The identities in my post show this is the case here. – Did Nov 8 '12 at 17:23
Okay, so you're suggesting that I use the fact that the identities I've stated hold path-wise (i.e. $\text{P-a.s.}$ under filtration equipped probability space with measure $P$), and therefore hold in all possible states of the world, and therefore the expressions are correct as they're stated. – Jase Nov 8 '12 at 17:27
One can forget the filtration context altogether. There are only functions, defined on some set $\Omega$. – Did Nov 8 '12 at 17:29
Your previous-to-last comment is correct (and one can omit almost surely in it). – Did Nov 9 '12 at 13:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.